Question: Divide the following rational expressions and simplify the result. $\dfrac{63n^2+168n}{56n-21} \div \dfrac{64n^2-9}{64n^2-48n+9}=$
Answer: Let's first factor the numerators and denominators of each expression separately. [Why are we doing this?] The numerator, $63n^2+168n$, of the first expression can be factored as $21n(3n+8)$ by factoring out $21n$. The denominator, $56n-21$, of the first expression can be factored as $7(8n-3)$ by factoring out $7$. The numerator, $64n^2-9$, of the second expression can be factored as $(8n+3)(8n-3)$ using the difference of squares pattern. The denominator, $64n^2-48n+9$, of the second expression can be factored as $(8n-3)(8n-3)$ using the perfect square pattern. Now the quotient looks as follows: $\dfrac{21n(3n+8)}{7(8n-3)} \div \dfrac{(8n+3)(8n-3)}{(8n-3)(8n-3)}$ To find the quotient of two rational expressions, we flip the divisor, multiply across, then simplify: [What's that?] $\begin{aligned} &\phantom{=}\dfrac{21n(3n+8)}{7(8n-3)} \div \dfrac{(8n+3)(8n-3)}{(8n-3)(8n-3)} \\\\\\ &= \dfrac{21n(3n+8)}{7(8n-3)} \cdot \dfrac{(8n-3)(8n-3)}{(8n+3)(8n-3)} &\text{Flip the divisor.}\\\\\\ &= \dfrac{21n(3n+8) \cdot (8n-3)(8n-3)}{7(8n-3) \cdot (8n+3)(8n-3)} &\text{Multiply across.}\\\\\\ &= \dfrac{{\cancel{7}}\cdot 3n(3n+8) {\cancel{(8n-3)}}{\cancel{(8n-3)}}}{{\cancel{7}}{\cancel{(8n-3)}} (8n+3){\cancel{(8n-3)}}} &\text{Cancel out common factors.}\\\\\\\\ &=\dfrac{3n(3n+8)}{8n+3} \end{aligned}$ Therefore, the simplified form of the quotient is $\dfrac{3n(3n+8)}{8n+3}$, which is equivalent to $\dfrac{9n^2+24n}{8n+3}$.